SOLUTION: During the first part of a trip, a canoiest travels 35 mls. at a certain speed. The canoiest travels 9 mls. on the second part of the trip. at a speed of 5 mls. slower. The total t

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: During the first part of a trip, a canoiest travels 35 mls. at a certain speed. The canoiest travels 9 mls. on the second part of the trip. at a speed of 5 mls. slower. The total t      Log On


   

Question 498174: During the first part of a trip, a canoiest travels 35 mls. at a certain speed. The canoiest travels 9 mls. on the second part of the trip. at a speed of 5 mls. slower. The total time for this trip is 2 hrs. What was the speed on each part of the trip?
Found 2 solutions by mananth, lwsshak3:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
first part 35 miles
second part 9 miles

speed in first part x mph
speed second part x -5 mph
Total rowing time 2 hours
Time first part 35 / x
Time second part 9 / ( x -5 )

Time first part + time second part = 2 hours

35 / x + 9 /(x -5 ) = 2
LCD = ( x + 0 )* (x -5 )
multiply the equation by the LCD
we get
35 * (x -5 )+ 9 x = 2
35 x -175 + 9 x = 2 X^2 + -10 x
54 x -175 = 2 X^2
2 X^2 -54 x 175 = 0
2 X^2+ -54 x+ 175 =
/ 2
1 X^2 -27 x 87.5 =

Find the roots of the equation by quadratic formula

a= 1 b= -27 c= 87.5

b^2-4ac= 729 - -350
b^2-4ac= 379 sqrt%28%09379%09%29= 19.47
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=( 27 + 19.47 )/ 2
x1= 23.23 mph
x2=( 27 -19.47 ) / 2
x2= 3.77 mph

First Part speed = 23.23 mph
Second part speed = 18.23 mph
m.ananth@hotmail.ca

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
During the first part of a trip, a canoiest travels 35 mls. at a certain speed. The canoiest travels 9 mls. on the second part of the trip. at a speed of 5 mls. slower. The total time for this trip is 2 hrs. What was the speed on each part of the trip?
**
What is m|s?
Assuming:
35mls=35 miles
9mls=9 miles
5mls=5 mph
..
let x=speed (mph) of first part of trip
x-5=speed (mph) of second part of trip
Travel time=distance/speed
35/x+9/(x-5)=2
LCD: x(x-5)
35(x-5)+9x=2x(x-5)
35x-175+9x=2x^2-10x
2x^2-54x+175=0
(let student solve for x with quadratic formula)
Answers are:
x=23.234 mph (reject, not reasonable)
or
x=3.766 mph